Find the degree 7 Taylor polynomial for
f(x )= 2cos(3x) centered at x = 3π/2
.
Thanks for your help
To find the degree 7 Taylor polynomial for f(x) = 2cos(3x) centered at x = 3π/2, we need to calculate its derivatives at x = 3π/2 up to the 7th derivative.
First, let's find the derivatives of f(x) = 2cos(3x):
f'(x) = -6sin(3x)
f''(x) = -18cos(3x)
f'''(x) = 54sin(3x)
f''''(x) = 162cos(3x)
f'''''(x) = -486sin(3x)
f''''''(x) = -1458cos(3x)
f'''''''(x) = 4374sin(3x)
Next, we can evaluate these derivatives at x = 3π/2:
f(3π/2) = 2cos(3(3π/2)) = 2cos(9π/2)
f'(3π/2) = -6sin(3(3π/2)) = -6sin(9π/2)
f''(3π/2) = -18cos(3(3π/2)) = -18cos(9π/2)
f'''(3π/2) = 54sin(3(3π/2)) = 54sin(9π/2)
f''''(3π/2) = 162cos(3(3π/2)) = 162cos(9π/2)
f'''''(3π/2) = -486sin(3(3π/2)) = -486sin(9π/2)
f''''''(3π/2) = -1458cos(3(3π/2)) = -1458cos(9π/2)
f'''''''(3π/2) = 4374sin(3(3π/2)) = 4374sin(9π/2)
Now, let's write out the Taylor polynomial:
P7(x) = f(3π/2) + f'(3π/2)(x - 3π/2) + f''(3π/2)(x - 3π/2)^2/2!
+ f'''(3π/2)(x - 3π/2)^3/3! + f''''(3π/2)(x - 3π/2)^4/4!
+ f'''''(3π/2)(x - 3π/2)^5/5! + f''''''(3π/2)(x - 3π/2)^6/6!
+ f'''''''(3π/2)(x - 3π/2)^7/7!
Substituting the values we calculated earlier, we get:
P7(x) = 2cos(9π/2) - 6sin(9π/2)(x - 3π/2) - 18cos(9π/2)(x - 3π/2)^2/2
+ 54sin(9π/2)(x - 3π/2)^3/6 + 162cos(9π/2)(x - 3π/2)^4/24
- 486sin(9π/2)(x - 3π/2)^5/120 - 1458cos(9π/2)(x - 3π/2)^6/720
+ 4374sin(9π/2)(x - 3π/2)^7/5040
Simplifying this expression gives the degree 7 Taylor polynomial for f(x) = 2cos(3x) centered at x = 3π/2.
To find the degree 7 Taylor polynomial for the function f(x) = 2cos(3x) centered at x = 3π/2, we need to find the values of the function and its derivatives at the center and use them to construct the polynomial.
First, let's find the derivatives of f(x) up to the 7th derivative:
f(x) = 2cos(3x)
f'(x) = -6sin(3x)
f''(x) = -18cos(3x)
f'''(x) = 54sin(3x)
f''''(x) = 162cos(3x)
f'''''(x) = -486sin(3x)
f''''''(x) = -1458cos(3x)
f'''''''(x) = 4374sin(3x)
Next, let's evaluate these derivatives at x = 3π/2:
f(x) = 2cos(3(3π/2)) = 2cos(9π/2) = 2cos((8π/2) + (π/2)) = 2cos(π/2) = 0
f'(x) = -6sin(3(3π/2)) = -6sin(9π/2) = -6sin((8π/2) + (π/2)) = -6sin(π/2) = -6
f''(x) = -18cos(3(3π/2)) = -18cos(9π/2) = -18cos((8π/2) + (π/2)) = -18cos(π/2) = 0
f'''(x) = 54sin(3(3π/2)) = 54sin(9π/2) = 54sin((8π/2) + (π/2)) = 54sin(π/2) = 54
f''''(x) = 162cos(3(3π/2)) = 162cos(9π/2) = 162cos((8π/2) + (π/2)) = 162cos(π/2) = 0
f'''''(x) = -486sin(3(3π/2)) = -486sin(9π/2) = -486sin((8π/2) + (π/2)) = -486sin(π/2) = -486
f''''''(x) = -1458cos(3(3π/2)) = -1458cos(9π/2) = -1458cos((8π/2) + (π/2)) = -1458cos(π/2) = 0
f'''''''(x) = 4374sin(3(3π/2)) = 4374sin(9π/2) = 4374sin((8π/2) + (π/2)) = 4374sin(π/2) = 4374
Now we can write the Taylor polynomial using these values:
f(x) = f(3π/2) + f'(3π/2)(x - 3π/2) + (f''(3π/2)/2!)(x - 3π/2)^2 + (f'''(3π/2)/3!)(x - 3π/2)^3 + (f''''(3π/2)/4!)(x - 3π/2)^4 + (f'''''(3π/2)/5!)(x - 3π/2)^5 + (f''''''(3π/2)/6!)(x - 3π/2)^6 + (f'''''''(3π/2)/7!)(x - 3π/2)^7
Simplifying this, since f(3π/2) = 0 and f''(3π/2) = 0 and f''''(3π/2) = 0, we get:
f(x) ≈ 0 + (-6)(x - 3π/2) + (0/2!)(x - 3π/2)^2 + (54/3!)(x - 3π/2)^3 + (0/4!)(x - 3π/2)^4 + (4374/5!)(x - 3π/2)^5 + (0/6!)(x - 3π/2)^6 + (0/7!)(x - 3π/2)^7
Simplifying further, we have:
f(x) ≈ -6(x - 3π/2) + (9/2)(x - 3π/2)^3 + (4374/120)(x - 3π/2)^5
So, the degree 7 Taylor polynomial for f(x) = 2cos(3x) centered at x = 3π/2 is:
f(x) ≈ -6(x - 3π/2) + (9/2)(x - 3π/2)^3 + (4374/120)(x - 3π/2)^5